wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A remote-sensing satellite of earth revolves in a circular orbit at a height of 0.25×106 m above the surface of earth. If earth's radius is 6.4×106 m and g=9.8 ms2, then the orbital speed of the satellite is

A
6.67kms1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
7.76kms1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
8.56kms1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
9.13kms1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 7.76kms1
The orbital velocity of the satellite is v=GMR+h where M= mass of earth and R= radius of earth.
so, v= 6.67×1011×(6×1024)(6.4×106)+(0.25×106)=7.76 km/s

flag
Suggest Corrections
thumbs-up
110
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon